5b^2+40b+35=0

Solution for 5b^2+40b+35=0 equation:

5b^2+40b+35=0

a = 5; b = 40; c = +35;
Δ = b2-4ac
Δ = 402-4·5·35
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-30}{2*5}=\frac{-70}{10}
=-7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+30}{2*5}=\frac{-10}{10}
=-1
$